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Sagot :
We can use the quadratic formula to solve this question. The quadratic formula is given below.
[tex]\begin{gathered} ax^2+bx+c=0 \\ \Rightarrow x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]In our case,
[tex]\begin{gathered} a=2,b=-2,c=-1 \\ \Rightarrow x=\frac{-(-2)\pm\sqrt[]{(-2)^2-4\cdot2\cdot-1}}{2\cdot2}=\frac{2\pm\sqrt[]{4+8}}{4}=\frac{2\pm\sqrt[]{12}}{4}=\frac{2\pm2\sqrt[]{3}}{4} \\ \Rightarrow x=\frac{1}{2}\pm\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]Then, the solutions are
[tex]x_1=\frac{1}{2}+\frac{\sqrt[]{3}}{2},x_2=\frac{1}{2}-\frac{\sqrt[]{3}}{2}[/tex]The solutions are 1/2+sqrt(3)/2 and 1/2-sqrt(3)/2
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